Q.Show that the points $A(2\hat{i}-\hat{j}+\hat{k})$, $B(\hat{i}-3\hat{j}-5\hat{k})$, $C(3\hat{i}-4\hat{j}-4\hat{k})$ are the vertices of a right angled triangle.

Showing the 12 most recent of 25 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If the equation of the plane passing through the points $(1,-3,2)$, $(-2,3,1)$ and perpendicular to the plane $x + 2y - 3z = 0$ is $ax + by + cz + d = 0$, then $\dfrac{a+b}{c+d} =$ (A) $\dfrac{3}{11}$ (B) $13$ (C) $\dfrac{13}{11}$ (D) $3$

   ›Reveal solutionSolution

   This tests finding a plane through two points and perpendicular to another plane, using the cross product of the connecting direction vector and the given plane's normal. Answer: $13$.

   Concept and Intuition

   A plane's normal vector must be perpendicular to every direction lying in the plane. Since the plane contains points $P_1(1,-3,2)$ and $P_2(-2,3,1)$, the vector $\vec{P_1P_2}$ lies in the plane, so the required normal $\vec n=(a,b,c)$ satisfies $\vec n\cdot\vec{P_1P_2}=0$. Also, "perpendicular to the plane $x+2y-3z=0$" means the two planes' normals are perpendicular, so $\vec n\cdot(1,2,-3)=0$. A vector perpendicular to both $\vec{P_1P_2}$ and $(1,2,-3)$ is simply their cross product.

   Step-by-Step Solution

   1. Direction vector: $\vec d = P_2-P_1 = (-2-1,\,3-(-3),\,1-2)=(-3,6,-1)$.
   2. Normal of given plane: $\vec n_2=(1,2,-3)$.
   3. Required normal: $\vec n=\vec d\times \vec n_2 = \begin{vmatrix}\vec i&\vec j&\vec k\\-3&6&-1\\1&2&-3\end{vmatrix}$ $= \vec i(6(-3)-(-1)(2)) - \vec j((-3)(-3)-(-1)(1)) + \vec k((-3)(2)-6(1))$ $= \vec i(-18+2) - \vec j(9+1) + \vec k(-6-6) = (-16,-10,-12)$. …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Let $\bar{a} = 3\bar{i} - \bar{j} - \bar{k}$, $\bar{b} = \bar{i} + \bar{j} - 2\bar{k}$ and $\bar{c} = 2\bar{i} + 2\bar{j} + \bar{k}$. Let $\bar{d}$ be a vector such that $|\bar{d}| = \sqrt{2}$ units. If the vector $\bar{d}$ is coplanar with $\bar{a}, \bar{b}$ and perpendicular to $\bar{c}$, then $\bar{d} =$ (A) $\pm\dfrac{1}{5}(3\bar{i} - 5\bar{j} + 4\bar{k})$ (B) $\pm\dfrac{1}{5}(-4\bar{i} + 5\bar{j} - 3\bar{k})$ (C) $\pm\dfrac{1}{5}(3\bar{i} + 5\bar{j} - 4\bar{k})$ (D) $\pm\dfrac{1}{5}(-3\bar{i} + 5\bar{j} + 4\bar{k})$

   ›Reveal solutionSolution

   $\bar d$ coplanar with $\bar a,\bar b$ means $\bar d=x\bar a+y\bar b$; perpendicularity to $\bar c$ fixes the ratio $x:y$; the given magnitude fixes the scale. The answer is (A).

   Concept and Intuition

   "Coplanar with $\bar a,\bar b$" means $\bar d$ lies in the plane spanned by $\bar a$ and $\bar b$, so it can be written as a linear combination $\bar d=x\bar a+y\bar b$ for some scalars $x,y$ (this is exactly what "coplanar with two given vectors, through the origin" means). The perpendicularity condition $\bar d\cdot\bar c=0$ then gives one constraint relating $x$ and $y$, so $\bar d$ is pinned down up to a single scalar multiple — which the given magnitude $|\bar d|=\sqrt2$ finally fixes (up to sign, since both directions along that line satisfy all the stated conditions).

   Step-by-Step Solution

   1. Given $\bar a=(3,-1,-1)$, $\bar b=(1,1,-2)$, $\bar c=(2,2,1)$.
   2. Since $\bar d$ is coplanar with $\bar a,\bar b$, write $\bar d = x\bar a + y\bar b = (3x+y,\,-x+y,\,-x-2y)$.
   3. Perpendicularity to $\bar c$: $\bar d\cdot\bar c=0$:

   $2(3x+y)+2(-x+y)+1(-x-2y)=0$

   $6x+2y-2x+2y-x-2y=0 \implies 3x+2y=0 \implies y=-\tfrac{3x}{2}.$

   4. Substitute back:

   $\bar d = \left(3x-\tfrac{3x}{2},\ -x-\tfrac{3x}{2},\ -x+3x\right) = \left(\tfrac{3x}{2},\,-\tfrac{5x}{2},\,2x\right).$

   Let $x=2t$ to clear fractions: $\bar d = (3t,-5t,4t) = t(3,-5,4)$. …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.In $\triangle ABC$, if $\overline{AB} = 2\bar{i} - \bar{j} + 2\bar{k}$ and $\overline{AC} = 3\bar{i} - 3\bar{j} + 4\bar{k}$, then that triangle $ABC$ is (A) an equilateral triangle (B) a right angled triangle (C) an isosceles triangle (D) a scalene triangle

   ›Reveal solutionSolution

   Compute all three side lengths of $\triangle ABC$ from $\overline{AB}$ and $\overline{AC}$ (with $\overline{BC}=\overline{AC}-\overline{AB}$) and classify by comparing them. Answer: isosceles triangle.

   Concept and Intuition

   Given two sides of a triangle as vectors from a common vertex, the third side is simply their difference (vector subtraction along the triangle). Once all three side lengths are known, classifying the triangle (scalene / isosceles / equilateral / right-angled) is a direct numeric comparison — no need for angles unless a right angle is suspected, in which case the converse of the Pythagorean theorem settles it.

   Step-by-Step Solution

   1. $\overline{AB}=(2,-1,2)$, so $|AB|=\sqrt{2^2+(-1)^2+2^2}=\sqrt{4+1+4}=\sqrt9=3$.
   2. $\overline{AC}=(3,-3,4)$, so $|AC|=\sqrt{9+9+16}=\sqrt{34}$.
   3. $\overline{BC}=\overline{AC}-\overline{AB}=(3-2,-3-(-1),4-2)=(1,-2,2)$, so $|BC|=\sqrt{1+4+4}=\sqrt9=3$.
   4. The three sides are $3, 3, \sqrt{34}$ — two equal sides, so the triangle is isosceles, not equilateral (all three would need to be equal) and not scalene (that needs all different). …

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4. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The vector of magnitude 2 lying in the plane of $\bar{a} = 2\bar{i} - \bar{j} + \bar{k}$ and $\bar{b} = \bar{i} + 3\bar{j} - 5\bar{k}$ and perpendicular to the vector $\bar{c} = \bar{i} + \bar{j} + \bar{k}$ is (A) $\dfrac{\sqrt{2}}{\sqrt{61}}(4\bar{i}+5\bar{j}-9\bar{k})$ (B) $\dfrac{\sqrt{2}}{\sqrt{9}}(2\bar{i}+3\bar{j}-5\bar{k})$ (C) $\dfrac{\sqrt{2}}{\sqrt{31}}(\bar{i}+5\bar{j}-6\bar{k})$ (D) $\dfrac{\sqrt{2}}{\sqrt{13}}(-\bar{i}-3\bar{j}+4\bar{k})$

   ›Reveal solutionSolution

   This tests writing a vector "in the plane of $\bar a,\bar b$" as a linear combination $\alpha\bar a+\beta\bar b$, using perpendicularity to $\bar c$ to pin the ratio $\alpha:\beta$, and finally scaling the resulting direction to the required magnitude.

   Concept and Intuition

   Every vector lying in the plane spanned by $\bar a$ and $\bar b$ is some linear combination $\alpha\bar a+\beta\bar b$ — that's what "lying in the plane" means. The extra condition (perpendicular to $\bar c$) gives one linear equation in $\alpha,\beta$, which fixes their ratio (the direction is determined up to an overall scale). The magnitude condition then fixes that scale.

   Step-by-Step Solution

   1. Let $\bar d=\alpha\bar a+\beta\bar b$ for some scalars $\alpha,\beta$ (this covers every vector in the plane of $\bar a,\bar b$).
   2. Require $\bar d\cdot\bar c=0$: $\alpha(\bar a\cdot\bar c)+\beta(\bar b\cdot\bar c)=0$.
   3. $\bar a\cdot\bar c=(2)(1)+(-1)(1)+(1)(1)=2-1+1=2$. $\bar b\cdot\bar c=(1)(1)+(3)(1)+(-5)(1)=1+3-5=-1$.
   4. So $2\alpha-\beta=0\Rightarrow\beta=2\alpha$. Taking $\alpha=1,\beta=2$: direction $=\bar a+2\bar b=(2+2,\,-1+6,\,1-10)=(4,5,-9)$.
   5. Magnitude of this direction: $\sqrt{4^2+5^2+(-9)^2}=\sqrt{16+25+81}=\sqrt{122}$. …

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5. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.Let the position vectors of the vertices of a triangle ABC be $\bar{a},\bar{b},\bar{c}$. If on the plane of the triangle, P is a point having position vector $\bar{x}$ such that $\bar{x}.(\bar{c}-\bar{b})=\bar{a}.\bar{c}-\bar{a}.\bar{b}$ and $\bar{x}.(\bar{a}-\bar{c})=\bar{a}.\bar{b}-\bar{b}.\bar{c}$, then for the triangle ABC, P is the (A) Centroid (B) Circumcentre (C) Incentre (D) Orthocentre

   ›Reveal solutionSolution

   Both given vector conditions are exactly the perpendicularity conditions defining two altitudes of the triangle, so P is the Orthocentre.

   Concept and Intuition

   A point lies on the altitude from vertex $A$ (dropped onto side $BC$) precisely when the vector from $A$ to that point is perpendicular to $BC$. Recognizing each given dot-product equation as such a perpendicularity condition (after regrouping terms) identifies which special triangle centre $P$ must be.

   Step-by-Step Solution

   1. First condition: $\bar{x}.(\bar{c}-\bar{b})=\bar{a}.\bar{c}-\bar{a}.\bar{b}$. The right side factors as $\bar{a}.(\bar{c}-\bar{b})$.
   2. So the condition becomes $\bar{x}.(\bar{c}-\bar{b}) - \bar{a}.(\bar{c}-\bar{b}) = 0$, i.e. $(\bar{x}-\bar{a}).(\bar{c}-\bar{b})=0$.
   3. This says the vector from $A$ (position $\bar{a}$) to $P$ (position $\bar{x}$) is perpendicular to $\bar{c}-\bar{b}$, which is the vector $\overrightarrow{BC}$. So $AP \perp BC$: P lies on the altitude from vertex A.
   4. Second condition: $\bar{x}.(\bar{a}-\bar{c})=\bar{a}.\bar{b}-\bar{b}.\bar{c}=\bar{b}.(\bar{a}-\bar{c})$.
   5. So $(\bar{x}-\bar{b}).(\bar{a}-\bar{c})=0$, meaning the vector from $B$ to $P$ is perpendicular to $\bar{a}-\bar{c}=\overrightarrow{CA}$. So $BP\perp CA$: P lies on the altitude from vertex B. …

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6. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.$\bar{a},\bar{b},\bar{c}$ are unit vectors. If $\bar{a},\bar{b}$ are perpendicular vectors, $(\bar{a}-\bar{c}).(\bar{b}+\bar{c})=0$ and $\bar{c}=l\bar{a}+m\bar{b}+n(\bar{a}\times\bar{b})$; (l, m, n are scalars), then $n^2=$ (A) $l^2+m^2$ (B) $-2lm$ (C) $2l-2m$ (D) $lm+l+m$

   ›Reveal solutionSolution

   Because $\bar a,\bar b,\bar a\times\bar b$ form an orthonormal triad, decomposing $\bar c$ in this basis and using the given perpendicularity condition shows $n^2=-2lm$.

   Concept and Intuition

   When $\bar a$ and $\bar b$ are perpendicular unit vectors, $\bar a\times\bar b$ is automatically a unit vector too (since $|\bar a\times\bar b|=|\bar a||\bar b|\sin 90°=1$) and is perpendicular to both $\bar a$ and $\bar b$. So $\{\bar a,\bar b,\bar a\times\bar b\}$ is an orthonormal basis — any vector's components along these three directions are just its dot products with each, and its squared magnitude is simply the sum of squared components (Pythagoras in 3D).

   Step-by-Step Solution

   1. Since $\bar a\perp\bar b$ and both are unit vectors, $\bar a.\bar b=0$ and $\{\bar a,\bar b,\bar a\times\bar b\}$ is orthonormal.
   2. Expand $(\bar a-\bar c).(\bar b+\bar c)=0$: $\bar a.\bar b+\bar a.\bar c-\bar c.\bar b-\bar c.\bar c=0$.
   3. Since $\bar a.\bar b=0$ and $\bar c.\bar c=|\bar c|^2=1$ (unit vector): $\bar a.\bar c-\bar b.\bar c-1=0 \Rightarrow \bar a.\bar c-\bar b.\bar c=1$.
   4. Given $\bar c=l\bar a+m\bar b+n(\bar a\times\bar b)$, dot with $\bar a$: $\bar a.\bar c=l(\bar a.\bar a)+m(\bar a.\bar b)+n\cdot\bar a.(\bar a\times\bar b)=l(1)+m(0)+n(0)=l$ (since $\bar a.(\bar a\times\bar b)=0$, a vector is always perpendicular to a cross product it's part of).
   5. Similarly, dot with $\bar b$: $\bar b.\bar c=l(\bar b.\bar a)+m(\bar b.\bar b)+n\cdot\bar b.(\bar a\times\bar b)=0+m(1)+0=m$.
   6. From step 3: $l-m=1$. …

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7. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.If the vectors $2\bar{i} + 3\bar{j} + l\bar{k}$, $-3\bar{i} - 2\bar{j} - 4l\bar{k}$ and $\bar{i} - \bar{j} + 3l\bar{k}$ form a right angled triangle for a positive value of $l$, then the length of its hypotenuse is (A) $\sqrt{\dfrac{40}{3}}$ (B) $\sqrt{\dfrac{55}{3}}$ (C) $\sqrt{\dfrac{65}{3}}$ (D) $\sqrt{\dfrac{59}{3}}$

   ›Reveal solutionSolution

   Because the three given vectors sum to zero, they are the side vectors of a closed triangle; finding which pair is mutually perpendicular locates the right angle, and the third side (opposite that angle) is the hypotenuse whose length we compute.

   Concept and Intuition

   If three vectors $u,v,w$ satisfy $u+v+w=\bar0$, they can be laid tip-to-tail to close a triangle — this is exactly the vector-polygon condition. The vertex where two of them (as drawn, not reversed) are mutually perpendicular is the right-angle vertex of the triangle, and the side "opposite" that vertex — i.e. the third vector — is the hypotenuse. So the whole problem reduces to (a) finding which pair dots to zero for some positive $l$, and (b) computing that third vector's magnitude.

   Step-by-Step Solution

   1. Let $u=(2,3,l)$, $v=(-3,-2,-4l)$, $w=(1,-1,3l)$.
   2. Check closure: $u+v+w=(2-3+1,\,3-2-1,\,l-4l+3l)=(0,0,0)$ — confirmed, they form a triangle.
   3. Test each pair's dot product for a value making it zero (this locates the right angle):
      • $u\cdot v = -6-6-4l^2=-12-4l^2$ — never zero for real $l$.
      • $v\cdot w = -3+2-12l^2=-1-12l^2$ — never zero for real $l$.
      • $u\cdot w = 2-3+3l^2=3l^2-1$ — zero when $l^2=\dfrac13$, i.e. $l=\dfrac{1}{\sqrt3}>0$. ✓ (matches "positive value of $l$" in the problem.) …

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8. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.A unit vector that is perpendicular to the vector $2\bar{i} - \bar{j} + 2\bar{k}$ and coplanar with the vectors $\bar{i} + \bar{j} - \bar{k}$ and $2\bar{i} + 2\bar{j} - \bar{k}$ is (A) $\dfrac{\bar{i} + 2\bar{j} + \bar{k}}{\sqrt{6}}$ (B) $\dfrac{3\bar{i} + 2\bar{j} - 2\bar{k}}{\sqrt{17}}$ (C) $\dfrac{2\bar{i} + 2\bar{j} - \bar{k}}{3}$ (D) $\dfrac{3\bar{i} + 2\bar{j} + 2\bar{k}}{\sqrt{17}}$

   ›Reveal solutionSolution

   Write the general coplanar combination of the two given vectors as $a\,p+b\,q$, impose perpendicularity to the third vector to pin down $a=0$, then normalize the resulting direction.

   Concept and Intuition

   "Coplanar with $p$ and $q$" means the target vector is some linear combination $a\,p+b\,q$ (this spans exactly the plane through the origin containing both). Imposing perpendicularity to a third given vector is then just one linear equation in $a,b$ — it typically forces a ratio (or here, forces one coefficient to vanish entirely), collapsing the family to a single direction, which we then normalize to a unit vector.

   Step-by-Step Solution

   1. General coplanar vector: $v=a(\bar i+\bar j-\bar k)+b(2\bar i+2\bar j-\bar k)=(a+2b)\bar i+(a+2b)\bar j+(-a-b)\bar k$.
   2. Require $v\perp(2\bar i-\bar j+2\bar k)$: $2(a+2b)-1(a+2b)+2(-a-b)=0$.
   3. Simplify: $(a+2b)(2-1)+2(-a-b)=(a+2b)-2a-2b=-a$.
   4. So the condition reduces to $-a=0\Rightarrow a=0$.
   5. With $a=0$: $v=b(2\bar i+2\bar j-\bar k)$, i.e. $v$ is parallel to $2\bar i+2\bar j-\bar k$.
   6. Magnitude of $2\bar i+2\bar j-\bar k$ is $\sqrt{4+4+1}=3$. …

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9. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If (2, -1, 3) is the foot of the perpendicular drawn from the origin (0, 0, 0) to a plane then the equation of that plane is (A) $2x + y - 3z + 6 = 0$ (B) $2x - y + 3z - 14 = 0$ (C) $2x - y + 3z - 13 = 0$ (D) $2x + y + 3z - 10 = 0$

   ›Reveal solutionSolution

   When the foot of the perpendicular from the origin to a plane is given, that

   point's position vector IS the plane's normal direction, and plugging the point

   back in gives the constant term instantly.

   Concept and Intuition

   The perpendicular from the origin to a plane is, by definition, along the

   plane's normal direction. So if $F=(x_0,y_0,z_0)$ is the foot of that

   perpendicular, the vector $\overrightarrow{OF}=(x_0,y_0,z_0)$ is normal to the plane,

   and the plane's equation is $x_0x+y_0y+z_0z = x_0^2+y_0^2+z_0^2$ (since $F$ itself

   must satisfy the plane equation, and $|OF|^2$ is exactly the dot product of $F$

   with itself).

   Step-by-Step Solution

   1. Normal direction $= (2,-1,3)$ (the given foot of perpendicular).
   2. Plane: $2x - y + 3z = k$ for some constant $k$.
   3. Since $(2,-1,3)$ lies on the plane: $k = 2(2) + (-1)(-1) + 3(3) = 4+1+9 = 14$. …

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10. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.The locus of a point at which the line joining the points $(-3, 1, 2), (1, -2, 4)$ subtends a right angle, is (A) $x^2 + y^2 + z^2 + 2x + y - 6z - 3 = 0$ (B) $x^2 + y^2 + z^2 + 2x - y - 6z + 3 = 0$ (C) $x^2 + y^2 + z^2 + 2x + y - 6z + 3 = 0$ (D) $x^2 + y^2 + z^2 - 2x + y - 6z + 3 = 0$

    ›Reveal solutionSolution

    This tests the classic 'locus subtending a right angle' problem, which is just the sphere having $AB$ as diameter, expressed via a perpendicularity dot-product condition. Answer: $x^2+y^2+z^2+2x+y-6z+3=0$.

    Concept and Intuition

    If a segment $AB$ subtends a right angle at a variable point $P$, then $PA\perp PB$, i.e. $\vec{PA}\cdot\vec{PB}=0$ for every such $P$ — this is exactly the defining property of a sphere with $AB$ as diameter (angle in a semicircle is a right angle, generalized to 3D). Writing this dot product in coordinates directly gives the sphere's equation.

    Step-by-Step Solution

    1. Let $P=(x,y,z)$. Then $\vec{PA}=A-P=(-3-x,\,1-y,\,2-z)$ and $\vec{PB}=B-P=(1-x,\,-2-y,\,4-z)$.
    2. Right angle at $P$: $\vec{PA}\cdot\vec{PB}=0$.
    3. $(-3-x)(1-x)=x^2+2x-3$ (expand: $-3+3x-x+x^2$).
    4. $(1-y)(-2-y)=y^2+y-2$ (expand: $-2-y+2y+y^2$).
    5. $(2-z)(4-z)=z^2-6z+8$ (expand: $8-2z-4z+z^2$). …

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11. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.Let ABC be an equilateral triangle of side $a$. M and N are two points on the sides AB and AC respectively such that $\overline{AN} = K\overline{AC}$ and $\overline{AB} = 3\overline{AM}$. If the vectors $\overline{BN}$ and $\overline{CM}$ are perpendicular, then $K =$ (A) $\dfrac{1}{5}$ (B) $\dfrac{2}{5}$ (C) $-\dfrac{1}{5}$ (D) $-\dfrac{2}{5}$

    ›Reveal solutionSolution

    Express $\overline{BN}$ and $\overline{CM}$ in terms of the two sides from $A$, use the $60^\circ$ dot product of an equilateral triangle, and set the perpendicularity condition to zero to solve for $K=\dfrac15$.

    Concept and Intuition

    Placing the vertex $A$ at the origin turns every other point into a simple scalar multiple of the two side vectors $\overline{AB}$ and $\overline{AC}$. Perpendicularity of two vectors becomes an algebraic condition: their dot product is zero. For an equilateral triangle, $\overline{AB}.\overline{AC}=a^2\cos 60^\circ=\dfrac{a^2}{2}$.

    Step-by-Step Solution

    1. Let $A$ be the origin, $\bar c=\overline{AB}$, $\bar b=\overline{AC}$, with $|\bar b|=|\bar c|=a$ and $\bar b.\bar c=\dfrac{a^2}{2}$.
    2. Since $\overline{AB}=3\overline{AM}$, $M=\dfrac{\bar c}{3}$. Since $\overline{AN}=K\overline{AC}$, $N=K\bar b$.
    3. $\overline{BN}=N-B=K\bar b-\bar c$, and $\overline{CM}=M-C=\dfrac{\bar c}{3}-\bar b$.
    4. Perpendicularity: $\overline{BN}.\overline{CM}=0$: $(K\bar b-\bar c).\left(\dfrac{\bar c}{3}-\bar b\right)= \dfrac{K}{3}(\bar b.\bar c) - K|\bar b|^2 - \dfrac13|\bar c|^2 + \bar b.\bar c = 0$.
    5. Substitute $|\bar b|^2=|\bar c|^2=a^2$, $\bar b.\bar c=a^2/2$: …

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12. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.$\bar{a} = \bar{i} - \bar{j} + \bar{k}$, $\bar{b} = 2\bar{i} + \bar{j} + \bar{k}$ are two vectors and $\bar{c}$ is a unit vector lying in the plane of $\bar{a}$ and $\bar{b}$. If $\bar{c}$ is perpendicular to $\bar{b}$ then $\bar{c}.(\bar{i}+\bar{j}+2\bar{k}) =$ (A) $0$ (B) $5$ (C) $\dfrac{1}{\sqrt{21}}$ (D) $\dfrac{2}{\sqrt{21}}$

    ›Reveal solutionSolution

    This tests finding a unit vector coplanar with two given vectors and perpendicular to one of them; the required dot product works out to $\dfrac{1}{\sqrt{21}}$.

    Concept and Intuition

    Any vector in the plane spanned by $\bar a$ and $\bar b$ can be written as a linear combination $m\bar a+n\bar b$. Imposing perpendicularity to $\bar b$ gives one linear equation in $m,n$, pinning down the direction of $\bar c$ up to a scalar (which is then fixed by the unit-length condition).

    Step-by-Step Solution

    1. Let $\bar c=m\bar a+n\bar b$ where $\bar a=(1,-1,1),\ \bar b=(2,1,1)$.
    2. $\bar c\cdot\bar b=0\Rightarrow m(\bar a\cdot\bar b)+n(\bar b\cdot\bar b)=0$.
    3. $\bar a\cdot\bar b=1(2)+(-1)(1)+1(1)=2-1+1=2$. $\bar b\cdot\bar b=4+1+1=6$.
    4. So $2m+6n=0\Rightarrow m=-3n$.
    5. $\bar c\parallel -3n\bar a+n\bar b=n(-3\bar a+\bar b)=n\big((-3,3,-3)+(2,1,1)\big)=n(-1,4,-2)$.
    6. Direction vector $(-1,4,-2)$ has magnitude $\sqrt{1+16+4}=\sqrt{21}$, so the unit vector is $\pm\dfrac{(-1,4,-2)}{\sqrt{21}}$. …

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